Answer:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F

F = 0.679 N
Answer:
The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.
Explanation:
Given that,
Mass of aircraft = 10000 kg
Speed = 620 km/h = 172.22 m/s
Altitude = 10 km = 1000 m
We calculate the change in potential energy





For g = 10 m/s²,
The change in potential energy will be 1000 MJ.
We calculate the change in kinetic energy





For g = 10 m/s²,
The change in kinetic energy will be 150 MJ.
Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.
Average speed is the answer
Answer:
a. 900 J
b. 0.383
Explanation:
According to the question, the given data is as follows
Horizontal force = 150 N
Packing crate = 40.0 kg
Distance = 6.00 m
Based on the above information
a. The work done by the 150-N force is


= 900 J
b. Now the coefficient of kinetic friction between the crate and surface is


= .383
We simply applied the above formulas so that each one part could calculate